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August 15th, 2018, 06:32 PM   #1
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Evaluating a limit

Hi,

Please evaluate the following limit:

$\displaystyle \lim_{x->0}\left ( \tfrac{1}{x^{2}}-\cot^{2}x \right )$

Thanks.

Last edited by skipjack; August 15th, 2018 at 07:41 PM.
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August 15th, 2018, 07:03 PM   #2
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The limit is 2/3.
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August 15th, 2018, 07:27 PM   #3
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Quote:
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The limit is 2/3.
Can you please explain?

Last edited by skipjack; August 15th, 2018 at 08:29 PM.
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August 15th, 2018, 07:44 PM   #4
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Assuming that the limit exists and that therefore $$\lim_{x \to 0} \left(\frac1{x^2}-\cot^2{x}\right) = L$$
setting $2t=x$ we have that
\begin{align}L &= \lim_{t \to 0} \left(\frac1{(2t)^2}-\cot^2{2t}\right) \\
&= \lim_{t \to 0} \left(\frac1{4t^2}-\frac{\cos^2{2t}}{\sin^2{2t}}\right) \\
&= \lim_{t \to 0} \left(\frac1{4t^2}-\frac{\cos^2{2t}}{(2\sin{t}\cos{t})^2}\right) \\
&= \lim_{t \to 0} \left(\frac1{4t^2}-\frac{(\cos^2{t}-\sin^2{t})^2}{4\sin^2{t}\cos^2{t}}\right) \\
4L &= \lim_{t \to 0} \left(\frac1{t^2}-\frac{\cos^4{t}-2\sin^2{t}\cos^2{t}+\sin^4{t}}{\sin^2{t}\cos^2{t}} \right) \\
3L &= \lim_{t \to 0} \left(\frac1{t^2}-\frac{\cos^4{t}-2\sin^2{t}\cos^2{t}+\sin^4{t}}{\sin^2{t}\cos^2{t}} \right) - \lim_{t \to 0} \left(\frac1{t^2}-\cot^2{t}\right) \\
&=\lim_{t \to 0} \left(\frac1{t^2}-\frac{\cos^4{t}-2\sin^2{t}\cos^2{t}+\sin^4{t}}{\sin^2{t}\cos^2{t}} - \frac1{t^2}+\cot^2{t}\right) \\
&=\lim_{t \to 0} \left(\frac{\cos^2{t}}{\sin^2{t}} - \frac{\cos^4{t}-2\sin^2{t}\cos^2{t}+\sin^4{t}}{\sin^2{t}\cos^2{t}} \right) \\
&= \lim_{t \to 0} \left(\frac{\cos^4{t}}{\sin^2{t}\cos^2{t}} - \frac{\cos^4{t}-2\sin^2{t}\cos^2{t}+\sin^4{t}}{\sin^2{t}\cos^2{t}} \right) \\
&= \lim_{t \to 0} \left(\frac{2\sin^2{t}\cos^2{t}+\sin^4{t}}{\sin^2{ t}\cos^2{t}} \right) \\
&= \lim_{t \to 0} \left(2+\tan^2{t} \right) \\
&=2 \\
L&=\frac23
\end{align}

Thanks to this for inspiration.
Thanks from happy21 and topsquark
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August 15th, 2018, 08:21 PM   #5
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Quote:
Originally Posted by v8archie View Post
Assuming that the limit exists and that therefore $$\lim_{x \to 0} \left(\frac1{x^2}-\cot^2{x}\right) = L$$
setting $2t=x$ we have that
\begin{align}L &= \lim_{t \to 0} \left(\frac1{(2t)^2}-\cot^2{2t}\right) \\
&= \lim_{t \to 0} \left(\frac1{4t^2}-\frac{\cos^2{2t}}{\sin^2{2t}}\right) \\
&= \lim_{t \to 0} \left(\frac1{4t^2}-\frac{\cos^2{2t}}{(2\sin{t}\cos{t})^2}\right) \\
&= \lim_{t \to 0} \left(\frac1{4t^2}-\frac{(\cos^2{t}-\sin^2{t})^2}{4\sin^2{t}\cos^2{t}}\right) \\
4L &= \lim_{t \to 0} \left(\frac1{t^2}-\frac{\cos^4{t}-2\sin^2{t}\cos^2{t}+\sin^4{t}}{\sin^2{t}\cos^2{t}} \right) \\
3L &= \lim_{t \to 0} \left(\frac1{t^2}-\frac{\cos^4{t}-2\sin^2{t}\cos^2{t}+\sin^4{t}}{\sin^2{t}\cos^2{t}} \right) - \lim_{t \to 0} \left(\frac1{t^2}-\cot^2{t}\right) \\
&=\lim_{t \to 0} \left(\frac1{t^2}-\frac{\cos^4{t}-2\sin^2{t}\cos^2{t}+\sin^4{t}}{\sin^2{t}\cos^2{t}} - \frac1{t^2}+\cot^2{t}\right) \\
&=\lim_{t \to 0} \left(\frac{\cos^2{t}}{\sin^2{t}} - \frac{\cos^4{t}-2\sin^2{t}\cos^2{t}+\sin^4{t}}{\sin^2{t}\cos^2{t}} \right) \\
&= \lim_{t \to 0} \left(\frac{\cos^4{t}}{\sin^2{t}\cos^2{t}} - \frac{\cos^4{t}-2\sin^2{t}\cos^2{t}+\sin^4{t}}{\sin^2{t}\cos^2{t}} \right) \\
&= \lim_{t \to 0} \left(\frac{2\sin^2{t}\cos^2{t}+\sin^4{t}}{\sin^2{ t}\cos^2{t}} \right) \\
&= \lim_{t \to 0} \left(2+\tan^2{t} \right) \\
&=2 \\
L&=\frac23
\end{align}

Thanks to this for inspiration.
Brilliant! This method is very nice though uncommon, I think!

Thx.
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August 15th, 2018, 09:06 PM   #6
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$\displaystyle \begin{align*}\cot^2\!x &= 1/(\sin^2\!x) - 1 \\
&= 1/(x^2(1 - x^2/3 \,+\; ...)) - 1 \\
&= (1/x^2)(1 + x^2/3 \,+\; ...) - 1 \\
&= 1/x^2 - 2/3 \,+\; ...\end{align*}$
so the required limit is 2/3.
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August 16th, 2018, 05:06 AM   #7
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Quote:
Originally Posted by happy21 View Post
Can you please explain?
If you show me some work I'll show you some work.
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August 16th, 2018, 09:59 AM   #8
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Quote:
Originally Posted by SDK View Post
If you show me some work I'll show you some work.
Yeah I did from the methods depicted above.
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